Question

Find the volume of the solid generated by revolving the region bounded by

y=5√sinx, y =0 and x1 = π/4 and x2 = π/2 about the x- axis.

Answer 1

`V =(25\pi)/(√(2))`

Explanation:

given,

y=5√sinx

Volume of the solid by revolving

`V = \int_a^b(\pi y^2)dx`

a and b are the limits of the integrals

now,

`V = \int_a^b(\pi (5√(sinx))^2)dx`

`V =25\pi \int_(\pi/4)^(\pi/2)sinxdx`

`\int sin x = - cos x`

`V =25\pi [-cos x]_(\pi/4)^(\pi/2)`

`V =25\pi [-cos (\pi/2)+cos(\pi/4)]`

`V =25\pi [0+(1)/(√(2))]`

`V =(25\pi)/(√(2))`

volume of the solid generated is equal to
`V =(25\pi)/(√(2))`